解下列方程组：
(1)$\begin{cases}x + y = 1, \\ 4x + y = 10;\end{cases}$ (2)$\begin{cases}x + 2y = 0, \\ 3x + 4y = 6;\end{cases}$
(3)$\begin{cases}3x - y = 7, \\ 2x + 3y = 1;\end{cases}$ (4)$\begin{cases}2a + 3b = 2, \\ 4a - 9b = -1;\end{cases}$
(5)$\begin{cases}x + 4y = 14, \\ \dfrac{x - 3}{4} - \dfrac{y - 3}{3} = \dfrac{1}{12};\end{cases}$ (6)$\begin{cases}3x - 4y = 1, \\ 2x + 3y = 12;\end{cases}$
(7)$\begin{cases}2018x + 2019y = 2020, \\ 2019x + 2018y = 2017;\end{cases}$ (8)$\dfrac{2m + 5n}{3} = \dfrac{m - 2n}{6} = 2$.
答案：(1)$\{ \begin{array} { l } { x = 3 }, \\ { y = - 2 } \end{array} $ (2)$\{ \begin{array} { l } { x = 6 }, \\ { y = - 3 } \end{array} $ (3)$\{ \begin{array} { l } { x = 2 }, \\ { y = - 1 } \end{array} $
(4)$\{ \begin{array} { l } { a = \frac { 1 } { 2 } }, \\ { b = \frac { 1 } { 3 } } \end{array} $ (5)$\{ \begin{array} { l } { x = 3 }, \\ { y = \frac { 11 } { 4 } } \end{array} $ (6)$\{ \begin{array} { l } { x = 3 }, \\ { y = 2 } \end{array} $
(7)$\{ \begin{array} { l } { x = - 1 }, \\ { y = 2 } \end{array} $ (8)$\{ \begin{array} { l } { m = 8 }, \\ { n = - 2 } \end{array} $
